What is the solution for y^{''}+4y^'+5y=0,y(pi/2)=0,y^'(pi/2)=2 ?

The solution for y^{''}+4y^'+5y=0,y(pi/2)=0,y^'(pi/2)=2 is y=-2e^{pi-2t}cos(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+5y=0,y(pi/2)=0,y^'(pi/2)=2

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 2

y = - 2 e^{π - 2 t} cos (t)

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = - 2 e^{π - 2 t} cos (t)

y = - 2 e^{π - 2 t} cos (t)

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 9

\frac{d ^{2} x}{d t ^{2}} + 58.49999999 x = 0

y^{^{''}} + 4 y^{^{'}} + 5 y = 0, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 7

y^{^{''}} - 4 y^{^{'}} - 21 y = 0, y (0) = a, y^{^{'}} (0) = 24

y^{^{''}} + 1 y = 0

What is the solution for y^{''}+4y^'+5y=0,y(pi/2)=0,y^'(pi/2)=2 ?
The solution for y^{''}+4y^'+5y=0,y(pi/2)=0,y^'(pi/2)=2 is y=-2e^{pi-2t}cos(t)